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I apologize for the simplistic questions. I have a retrieval process that has a set of random and systematic uncertainties associated with it. I'm assuming that these are all independent. The goal is to find the total error for this process. I have two questions related to this:

1) To find the total systematic uncertainty can I sum all the systematic uncertainty components in quadrature? Assuming that they are independent, it is unlikely that they will all contribute in the same direction and it seems to make sense to add them in quadrature. However, by definition they are not random. Does this mean they must be added linearly?

2) To find the total uncertainty (total_random + total_systematic), can the total_random and total_systematic be added in quadrature or must they be added linearly?

Thank you in advance for you help!

user8675309
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  • What are the uncertainties? Are they standard deviations, or variances, or something else? – TooTone Mar 07 '14 at 17:15
  • ok, just to check, variances mean that the units of the uncertainties is the square of the scale of the underlying measurement. The units of standard deviation are the same as the units of the underlying measurement and it also makes sense with standard deviation to talk about adding in quadrature – TooTone Mar 07 '14 at 17:31
  • @TooTone The variance of the uncertainties are added together to get the total uncertainty and then the square-root is taken for reporting (std) for reporting the value. So even though the we are dealing with the std (or variance) of a systematic bias it makes sense to use quadrature? Thanks again! – user8675309 Mar 07 '14 at 17:50
  • Sorry I missed that but your comment helps. I was in the middle of writing an answer, please take a look. In general, I agree it makes sense to have the system work internally in terms of variances and report standard deviations. – TooTone Mar 07 '14 at 17:58
  • In my answer I've made (at least) one big assumption, see the discussion under the answer. Please can you comment as to whether I've understood you properly (here or under the answer). – TooTone Mar 07 '14 at 18:17

1 Answers1

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1) To find the total systematic uncertainty can I sum all the systematic uncertainty components in quadrature? Assuming that they are independent, it is unlikely that they will all contribute in the same direction and it seems to make sense to add them in quadrature. However, by definition they are not random. Does this mean they must be added linearly?

I think what you mean here is that all the systematic uncertainty components are functions of the same underlying systematic uncertainty. Mathematically, there is some underlying systematic uncertainty random variable $S$, and each systematic component is some constant, or weight, $s_i$ times $S$. The $i$th system component can then be expressed as follows. $$ S_i = s_iS$$

Without loss of generality, let the variance of $S$ be 1. If the variance of each systematic component is $v_i$, then $s_i=\sqrt{v_i}$, and the total systematic variance is given by \begin{align} \text{Var}\left(\sum_i S_i\right) &= \text{Var}\left( \sum s_i S\right)\\ &= \left( \sum_i s_i \right)^2 \text{Var}(S)\\ &= \left( \sum_i s_i \right)^2\\ &= \left( \sum_i \sqrt{v_i} \right)^2 \end{align}

If your uncertainties had been specified in terms of standard deviations, then $s_i$ would be the standard deviation of each component and the answer would be simply be $\sum_i s_i$, i.e. they must be added linearly.

2) To find the total uncertainty (total_random + total_systematic), can the total_random and total_systematic be added in quadrature or must they be added linearly?

They are independent, and for any two random variables $X,Y$ that are independent, the variance of the sum is the sum of the variances, i.e. $\text{Var}(X+Y)=\text{Var}(X) + \text{Var}(Y)$. So if the total random and total system uncertainties are given as variances, you simply add them together. (Assuming that individual random uncertainties are independent of one another, the same applies to forming the total random uncertainty from individual random uncertainties.)

If your uncertainties are specified in terms of standard deviations, then you need to add in quadrature.

TooTone
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  • I was initially confused by this answer. What got me was the assumption that systematic uncertainty can be modeled as proportions of a single random variable. How does it even make sense to model *systematic* uncertainty as random? What exactly is your distinction between "systematic" and "random" uncertainties? What justifies decomposing *all* "systematic" uncertainties as proportions of a single variable $S$? Perhaps a concrete example would clarify your model. – whuber Mar 07 '14 at 18:09
  • @whuber thanks for your comment, I can see where you're coming from. I tried to reason as to what the question was saying based on the fact that if the systematic components were all independent then they might as well all be random. My answer was also based on models I've come across in the past. I may be wrong -- the assumption of a single systematic component is the big one, I agree. An example of such a model is CAPM, where every stock in a market has a systematic market component and a random residual component. I've also seen this extended to multiple systematic components per stock. – TooTone Mar 07 '14 at 18:14
  • I believe there will be some scope for your answer to be correct, because it does seem to capture some of the essence of a difference between different kinds of variation, but I'm not yet sure what that scope is. That's why I was asking for a clarification of your meaning and perhaps an example--not to prove you wrong, but to help elucidate the circumstances under which you are right. – whuber Mar 07 '14 at 18:20
  • @whuber no worries I welcome all your comments. (I still remember one great comment you made some time ago which pointed out an absolute stinker of a mistake on a post I made. I can laugh about it now:) In this case, in hindsight it would have been better for me to have written "I'm going to assume that" rather than "I think what you mean here is that". Here I felt like I had a good idea what the OP was after, but I may have been wrong, in which case it would have served me right for not clarifying further before answering. – TooTone Mar 07 '14 at 18:23
  • @TooTone thanks for the explanation. I'm trying to understand the difference between the answer for the two questions. For the first question you have the STD summed linearly for the second question you say in quadrature. Does this stem from the fact that in the first question the variances are components of a single systematic uncertainty? – user8675309 Mar 07 '14 at 18:31
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    @user877334 yes, exactly, but if you look at the discussion above, you will want to be sure that the model I'm using is the same as the one that you are using! – TooTone Mar 07 '14 at 18:33
  • also, thanks for accepting my answer. Note if you get another one you like better later on, you can unaccept mine and accept the other one instead (for this reason questioners will sometimes hold off accepting an answer for some hours. Also, when you have more reputation you will be able to upvote before accepting.). – TooTone Mar 07 '14 at 18:38